Irreducible components of a affine cone are cones Let $\theta:A^{n+1}-\{(0,\dots,0)\}\to P^n$. Define cone above $Y\subset P^n$ as $C(Y)=\theta^{-1}(Y)\cup\{(0,\dots,0)\}\subset A^{n+1}$. It is clear that there is a surjective map by restricting to $C(Y)$ by $\theta\vert_{C(Y)-\{(0,\dots,0)}:C(Y)-\{(0,\dots,0)\to Y$. Let $I_p(Y)\subset S$ where $S$ is the graded ring and $I_p(Y)$ is the ideal corresponding to projective variety. It is clear that $I(C(Y))=I_p(Y)$ by homogeniety of polynomials. 


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*What is the topology on this $C(Y)$ here? Do I use the topology induced by $P^n$ such that $\theta$ is continuous? If that is the case, then I must have any component of $C(Y)$ must be a cone. It seems to be the case as in the post Every irreducible component of an affine cone contains its vertex.

*If that is not the case, I do not see the reason that irreducible component of $C(Y)$ has to be a cone say taking a point $p\in C(Y)$. Why should it be a cone then from geometric viewpoint or algebraic viewpoint please? I can understand the proof in the other post but I do not feel confident that I fully understood the whole picture as I do not know what kind of topology is on $C(Y)$ here or even $A^{n+1}-\{(0,\dots,0)\}$. 

I will explain a bit on why I am asking this question. I want to show $C(Y)$ is irreducible if and only if $Y$ is irreducible from purely topological argument. As I expect that irreducibility should be independent of algebraic construction, "Only If" side is trivially done by topological argument. 
If $Y$ is irreducible, $C(Y)$ is irreducible. Assume $C(Y)$ is reducible. Then $C(Y)=C_1\cup C_2$. So there is $fg=0$ on $C(Y)$. Thus $fg\in I(C(Y))$ which is homogeneous. Thus it suffices to assume $fg$ homogeneous by dilation symmetry(i.e. $f(kx)=k^{deg(f)}f(x)$). Hence $f,g$ are homogeneous factors of a homogeneous polynomial. Thus $C_1,C_2$ are cones which corresponds to some variety of $Y$ and their image union is exactly $Y$. This is contradiction. 
The "If"$ direction requires some algebraic manipulation which should be avoided. 


*Can I avoid quoting algebraic result to prove the "If" direction?

 A: First note that $I(Y)=I(C(Y))\subset k[x_0,\dots , x_n]$, and $Y$ is rreducible $\iff$ $I(Y)$ is prime $\iff$ $C(Y)$ is irreducible. 


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*The topology is the subspace topology induced from the Zariski topology on $A^{n+1}$.

*In fact  if $Y=Y_1\cup \cdots \cup Y_r$ with all $Y_r$ irreducible, then $Y=Y_1\cup \cdots \cup Y_r$, then $C(Y)=C(Y_1)\cup \cdots \cup C(Y_r)$ with all $C(Y_i)$ irreducible, Because both of them correspond to the primary decomposition $I(Y)=I(Y_1)\cap \cdots \cap I(Y_r)$.

*It is possible to prove "If" direction more geometrically (when $k$ is infinite).
Suppose $Y$ is irreducible and $X=C(Y)$, $\pi:X \to Y$ the projection map. For each $y\in Y$, let us denote the the fibre $\pi^{-1}(y)$ by $X_y$.
Now suppose we have $X=X_1 \cup X_2$ as a union of two closed subsets, then for each $y\in Y$. $X_y=(X_1\cap X_y) \cup (X_2 \cap X_2)$, so $X_y\subset X_1$ or $X_y \subset X_2$ for each y. Let $Z_i=\{ y\in Y| X_y\subset X_i \}$, it suffices to show that $Z_1=Y$ or $Z_2=Y$. But since $Y$ is irreducible ,it suffices to show that $Z_i$'s are closed. To prove this, note that $Z_i=\pi (\tilde{X_i})$, where $\tilde{X_i}= V(\{ f(tx_0,\dots ,tx_n)| f\in I(X_i),  t\in k  \})$. It can be seen that $\tilde{X_i}$ is a cone (i.e. $I(\tilde{X_i})$ a homogeneous ideal, here I assume $k$ is infinite). Therefore $Z_i= V(I(\tilde{X_i}))$ is closed.
The statement is not true when $k$ is finite, a subset of $A^n$(or $P^n$) is irreducible if and only if it is a singleton in this case. 
Also, the proof does use some algebra and I don't think there is a purely geometric proof for the "If" part since the definition of $C(X)$ and Zariski topology are algebraic. In fact, one can consider the map $A^1\sqcup \{ \text{point} \} \to P^1$, which is a surjective map to an irreducible space with irreducible fibers, but the source is not irreducible. However, if $\pi : X \to Y$ is a closed surjective such that all fibers are irreducible of the same dimension and $Y$ is irreducible, then we can conclude $X$ is irreducible as in this post Use irreducible fibers to show $X$ is irreducible.
